SOLUTION: The cross section of a trough is an isosceles trapezoid. If the trough is made by bending up the sides of a strip of metal C inches wide, what should be the angle of inclination of
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Question 1083631: The cross section of a trough is an isosceles trapezoid. If the trough is made by bending up the sides of a strip of metal C inches wide, what should be the angle of inclination of the sides and the width across the bottom if the cross-sectional area is to be a maximum?
Answer: Angle= ;
width across the bottom is ;
maximum area is
Thank you, I promise work in some problems coming soon. Answer by htmentor(1343) (Show Source):
You can put this solution on YOUR website!
The area of the trapezoid pictured above is H(W+X)
We need to express this in terms of the angle a and the width W
From the figure we see that H = Lcos(a) and X = Lsin(a)
Since C = W + 2L we have L = (C-W)/2
Thus H = 1/2(C-W)cos(a) and X = 1/2(C-W)sin(a)
Hence the area A = (W/2)(C-W)*cos(a)+((1/2)(C-W))^2*sin(a)*cos(a)
To maximize the area in terms of both W and a, we take the partial derivatives
w.r.t. each variable and set equal to zero.
First we maximize in terms of W:
dA/dW = 0 = (C/2 - W)*cos(a) + (1/4)sin(a)*cos(a)(2W - 2C)
(C/2 - W)*cos(a) - (1/2)(C - W)sin(a)*cos(a) = 0
Multiply through by 2 and divide through by cos(a):
C - 2W - (C - W)sin(a) = 0
Solve for sin(a):
sin(a) = (C - 2W)/(C - W) [1]
Now maximize the area in terms of a:
dA/da = 0 = (W/2)(C-W)(-sin(a)) + (1/2)(C-W)^2(cos^2(a)-sin^2(a))
Using the identity cos^2(a) = 1 - sin^2(a), we can write
(-W/2)(C-W)sin(a) + (1/2)(C-W)^2(1 - 2sin^2(a)) = 0
Divide through by (C-W)/2:
-W*sin(a) + (C-W)(1 - 2sin^2(a)) = 0
Substitute the value for sin(a) in [1] above:
-W*(2W-C)/(W-C) + (C-W)(1-2((2W-C)/(W-C))^2)
Carrying out the algebra and collecting like terms we are left with:
2W - C - (W-C)/2 = 0
2W - W/2 - C/2 -> W = C/3
Use this expression to determine the value of the angle a:
sin(a) = (2(C/3) - C)/(C/3 - C) = 1/2 -> a = pi/6
This means that the inclination angle from the horizontal is pi/2 - pi/6 = pi/3
Substituting the values for sin(a), cos(a) and W, we can solve for the area in terms of C:
A = (C/3)((C-(C/3))/2)*sqrt(3)/2 + (1/2)((C-C/3)/2)^2*sqrt(3)/2
This gives A = C^2/(4*sqrt(3))
